GENERAL CHEMISTRY SECTION II: CHEMICAL BONDING

LECTURE 7: AN INTRODUCTION TO THE CHEMICAL BOND

Chapter Summary:

• We use the concept of valence electrons surrounding atoms to create 2-dimensional structures in which species with chemical bonds are more stable than solitary atoms.

• We base the stability argument for chemical bonds on the generation of species that have achieved stable noble gas configurations (s2p6).

• A first kind of bond is ionic: electrons are donated or accepted to create anions and cations that have s2p6 configurations.

• A second kind of bond is covalent: these occur when non-metal atoms share electrons with one another. • Many compounds achieve s2p6 configurations by sharing electrons in overlapping p orbits to form double and triple

bonds as well as resonance structures. • Bonds can form even when s2p6 configurations are not possible – for example, atoms like boron have 3 valence

electrons and usually can form only 3 covalent bonds, with the final molecular bonds having just 6 electrons. Compounds that contain an odd number of electrons form radicals that can’t satisfy the octet rule. Larger atoms (≥ n=3) have available d orbits, able to hold five and six bonds.

• Many compounds can have multiple Lewis structures that satisfy the octet rule. It’s necessary to evaluate the formal charge of atoms in the resonance structure to find the lowest energy, most stable structures.

• Electronegativity is a measure of the pull that an atom has on electrons in a chemical bond. Calculation of the difference in EN assists in distinguishing ionic and covalent character of bonds.

• Trends in bond energy and bond length can be predicted. Bonds with multiple charges are higher energy and shorter.

110

110

BEFORE WE TAKE ON BONDING: SOME FINAL PERIODIC TABLE TRENDS As we leave atoms and take on bonding, we need to look at the properties of the main group elements, which are those with valence electrons in s and p orbitals:

These main group materials exhibit very distinct changes by group. This makes for easy chemistry to teach. Trends are actually present. Speaking of trends, a new one, metallic character, is in the main group elements. The entire left side of the table is metallic, with Cs being the most metallic. The right side is more nonmetallic, but as you go down the table in each group, metallic character increases. There are also some metalloids in the p-block of the table; these are nonmetallic elements with metallic properties. Examples:

• O2 and S2 are nonmetal gases. • Te and Po are metalloids.

111

111

TRANSITION METALS

Let’s be honest – d-block elements are a mess. There are no trends, properties are similar, and they all look like metals. All of this makes them awful to teach in this course.

Why are transition metals so similar? – The electrons in d orbitals are inner electrons (i.e., not valence; recall: one d orbital has lower energy than the next s orbital). Even still, when a transition metal’s number of e- changes, it’s the d orbital electrons that are being changed. So among the transition metals, valence e- are pretty much constant, and properties are the same. Can this be a good thing? Well, their similar atomic radii let us mix-and-match them to form alloys with neat properties, like superconductors. Plus, lots of oxidation numbers mean lots of e- transfer in biological systems.

NOW TO BONDING: A COMPARISON OF COVALENT AND IONIC BONDS

Moving on to the big picture! We create 2D Lewis structures by drawing structures of atoms with their valence electrons assigned:

And mixing and matching to form chemical species in which the duplet or octet is satisfied.

A bond is what forms when two electrons are placed in the same orbital and the overall energy is lowered. There are two kinds of bonds: ionic and covalent bonds.

Notice that the numbers running along the top are the number of valence electrons.

112

112

Covalent bonds involve electrons being shared between two atoms.

Ionic bonds involve a donation of electrons from one atom to another.

ANOTHER PERIODIC TREND: IONIC RADII

Just like the trends we’ve learned about in previous lectures, effective nuclear charge explains ionic radii. But there’s a complication, because the trend doesn’t follow the periodic table, which was built around neutral atoms (with equal numbers of protons and electrons). So to see a smooth trend in ENC – and therefore in ionic radii – we need to rearrange the table in terms of isoelectronic species: atoms or ions with the same number of electrons (which also means they have the same electronic configurations). Since isoelectronic species have a certain number of e-, there is obviously one element with that number of e- in its neutral form. But the group also has ions that have ended up with that same number, after their neutral form lost or gained e-. Here’s an example where each atom or ion has 10 e-:

N3- O2- F1- Ne Na1+ Mg2+ Al3+ e- 10 10 10 10 10 10 10 p+ 7 8 9 10 11 12 13

Smallest ENC Largest ionic radii

Largest ENC Smallest ionic radii

113

113

Now, since atoms/ions in an isoelectronic species have the same number of electrons but different numbers of protons, each will have its own ENC and, consequently, radius size. And there’s a trend to these ionic radii – here are some ions from that table, lined up by relative size:

Notice that if you were to put the ions into their respective groups, the order would be: 16 ! 17 ! 18 (noble gases) ! 1 ! 2. This mirrors the argument I make in class that an easy way to create an ordered isoelectronic table is to tear the periodic table in half – as in, cut it down the middle and switch the sides, so that group 1 ends up sitting to the right of the noble gases. We also need to move the new right-hand half up by one row, ending up with a table that has Ne (n = 2) and Na (n = 3) side by side. Here’s a picture to help you visualize:

The point of this is to see the ionic radii trend as it goes along the periodic table, so that the ions are ordered by size. It’s important to notice that each of the last three graphics shows the exact same ions in the exact same order. It’s just that, for that final one, we needed to rearrange the periodic table a little to make that order happen.

114

114

THEORY OF IONIC BONDS Ionic bond formation is often dependent on the phases of the molecules. For example, if we wanted to know whether

Na(g) + Cl(g) ! Na+ and Cl-

is energetically favored, we could calculate the difference in ionization energy of Na(g) and electron affinity of Cl(g):

Since Na(g) + Cl(g) ! Na+(g) + Cl-

(g) requires energy to happen, it’s not energetically favored. But what if our product was a salt instead of gas? This means there’d be energy – called lattice energy – holding the ions together, due to Coulombic attraction:

So our favored reaction turns out to be: Na(g) + Cl(g) ! NaCl(s)

115

115

The important thing to note is that even though ion formation is not favored in the gas phase, it is favored as a solid because of all the Coulombic attraction present in salt crystal.

What kind of ionic compounds form? Look at periodic table to see what combinations of electron-donating and electron-accepting groups can match up – and with what stoichiometry they do – to achieve a neutral salt.

There are nine possible forms of ionic salts:

How do we draw structures of ionic bonds? The general Lewis structure of ions is:

We want to find combinations of X and Y that yield a molecule with 8 electrons.

116

116

TIME OUT FOR AN ASIDE ON LATTICE ENERGY We can develop a reasonable argument with charge density considerations that explains the magnitude of lattice energy. Consider four ions:

When we take ions like this and combine them with anions, the higher the charge density, the closer the bonds, and the higher the crystal lattice energy.

Example: Rank in decreasing order the lattice energies of NaF, MgCl2, KCl. Solution: MgCl2 > NaF > KCl MgCl2 has the highest charge density because of its multiply-charged ion (Mg2-). NaF and KCl each have only singly-charged ions, and since NaF is smaller than KCl, KCl has the lower charge density. (Hint: for knowing molecules’ size, remember the atomic radius trend.)

117

117

LECTURE 8: INTRODUCTION TO COVALENT LEWIS DOT STRUCTURES In the upper right corner of the periodic table are the nonmetals. These relatively few atoms have the following Lewis structures:

And we still want to mix and match them to create filled shells, like s2 and s2p6. Here are some examples:

In all cases, covalent bonds involve the sharing of electrons. No ions are formed during covalent bonding. But there’s an important “asterisk” to covalent bonds: though the electrons are shared, they’re not necessarily shared equally – one atom may tend to “hog” the e-’s and keep them on its side of the bond. Electronegativity will explain this.

118

118

A SYSTEMATIC APPROACH TO DRAWING LEWIS STRUCTURES Is there a systematic way to build Lewis structures of molecules? Try this:

1. Write out the atoms symmetrically, with the most electropositive in the center:

2. Count up the total number of valence electrons (e.g., hydrogen = 1, carbon = 4, nitrogen = 5):

CH4 = 4 + 4(1) = 8 NCl3 = 5 + 3(7) = 26 C2H4 = 2(4) + 4(1) = 12

3. Create octets and duplets for perimeter atoms:

4. Put the remaining e-’s around the central atom:

119

119

Any other helpful hints? Yes – there are a limited number of shapes based upon spreading e- pairs out around the central atom. So know that you need to have an open mind about possible expanded sizes to the Lewis structures.

Any other help? Yes – there is a simple calculation involving the number of shared electrons and number of bonding sites that tells you what kind of bonding (i.e., single, double, triple) is there:

The white triangles ( ) are bonds that are also on the z-axis, and they are pointing out of the front of the page. (Like if someone did the same nail/hammer thing (see other text box), except from the back of the page; you’d be looking at the pointy end and the nail shaft would be coming out at you on this side of the page, just like one of these bonds.)

The striped triangles ( ) are bonds on the z-axis (perpendicular to the page); they’re pointing away from you, into the page. (Like if you put the pointy end of a nail to this smiley face ! and hammered it through the page; the shaft of the nail would be oriented like one of these bonds.)

120

120

This can be very helpful when it works (when the octet rule is satisfied) but really misleading when it doesn’t (when you have hypo- or hypervalent structures that don’t satisfy the octet rule). I personally never use it, but, then again, I am a chemistry professor. You might find it a good way to get started on learning this material when you are in doubt about resonance or the number of double or triple bonds.

EXPLAINING RESONANCE Drawing Lewis structures lets us explain resonance, which is a property of a molecule whose structure has multiple, equivalent ways of being drawn. When this happens, we can draw multiple Lewis structures. The examples we see most often are when 3 e-’s are placed in 2 regions of e-’s:

Note that when we get lazy, we draw… …instead of all the different structures:

121

121

HYPERVALENT COMPOUNDS Hypervalent compounds are ones with 5 or 6 electron-rich regions. They do exist, and they’re nothing to panic about. There are only 2 possible Lewis structures for these:

Example – PCl5 has 40 valence e-’s to distribute in its Lewis structure.

• This structure shows us that P must accommodate 5 bonds. Example – XeF4 has 36 valence e-’s to distribute in its Lewis structure.

• After we put 32 electrons around the perimeter atoms (the F’s), we still had 4 e-’s to add to our compound. Those were added to the central atom, so it ended up with 6 electron-rich regions – 4 of them single bonds between Xe and F, and 2 of them non-bonding electrons (lone pairs) on Xe.

122

122

MULTIPLE CENTRAL ATOMS I’m bored by all these single central atoms. What about molecules with multiple central atoms? How do we draw Lewis structures for those? Well, we draw them just as we drew molecules with a single central atom:

1. Lay out all the atoms. 2. Count the total number of valence electrons. 3. Give the perimeter atoms their valence e-’s. 4. Distribute the remaining electrons around the center atoms.

Here are 3 famous examples:

You’ll do this for all organic compounds. They almost always end up with 8 e-’s around each carbon (as in, its n = 2 shell gets filled) and 2 on each hydrogen (its n = 1 shell gets filled).

123

123

IS THE OCTET/DUPLET RULE ALWAYS FOLLOWED? A preface in the form of a high school chemistry flashback:

The octet/duplet rule says that atoms will covalently bond to form molecules in which their outermost/valence shells are filled. So since n = 2 atoms’ outermost shell can hold up to 8 electrons, they bond in such a way that gives that shell an octet – a set of 8 electrons – because a full valence shell means a stable configuration. The same idea applies to the n = 1 atoms (H and He), except that their outermost shell hold up to 2 electrons, so a full shell has a duplet.

Now, back to the question/section title. The answer is no – there are 3 exceptions to the octet/duplet rule:

• The hypervalent case – more than 8 electrons are around the central atom, like in the cases of SF6, XeF4, IBr2

-, PCl5, etc. This happens with n ≥ 3 atoms, because they have available d orbitals.

• The hypovalent case – fewer than 8 electrons are around the central atom, like in molecules with B or

Be as the central atom. There are too few electrons to form an octet. • H has 1 valence e- " it can use it to form 1 bond. • Be has 2 valence e-’s " it can use them to form 2 bonds. • B has 3 valence e-’s" it can use them to form 3 bonds.

• The odd case – valence shells with an odd number of electrons – not all of the electrons can “buddy up,” so one electron pair can’t form. This means that, instead of an “electron pair,” there will be one radical.

Hint: if you ever see a compound with boron, it will be a hypovalent exception to the octet rule.

124

124

LECTURE 9: MORE SOPHISTICATED IDEAS IN LEWIS DOT STRUCTURES Can we make things more difficult? Sure!

FORMAL CHARGE Let’s discuss formal charge. First, why do it? Consider the following examples of two molecules – CH2O and SCN- – that each have multiple Lewis structures fitting our rules:

We need to find the formal charge of each atom to solve this problem. The compound with the smallest formal charge values is the most stable, which makes it the correct Lewis structure for that molecule. So how do you calculate FC?

Formal charge = (# of the atom’s valence e-’s) - (2e-’s for every unbonded e--pair + 1e- from each bonded e--pair)

125

125

Let’s assign the formal charges for the atoms in CH4:

Example:

Which of the following Lewis structures for CH2O is more stable?

Solution: Example:

Which of the following Lewis structures for SCN- is most stable? Solution:

126

126

ELECTRONEGATIVITY How do we distinguish whether a bond is ionic or covalent? We can determine this in a quantitative manner using electronegativity:

∆ EN < 1.5 is covalent ∆ EN > 1.5 is ionic

But what is EN? Electronegativity is a measure of an atom’s pull on electrons. For the figures below, the atoms on the right side of the bond pull more (have higher EN), and the atoms on the left pull less (have lower EN).

So where do these numbers for ∆EN come from? – Both Linus Pauling and Robert Millikan made scales. You’ll have to know some elements’ EN values (mostly from the second row), but they’re easy – here is a rough version that makes them easy to remember in calculations. Starting on the left of the second row, assign Li = 1.0, and add 0.5 to each successive element as you move to the right across the table; you should end up with F = 4.0. Another (final) element to know is H, which is H = 2.2. And that’s all there is to it!

127

127

Just like other trends, this EN trend can be drawn as a diagonal from the bottom left of the table to the top right. This is also due to effective nuclear charge. Example:

What are the |ΔEN| values for CO, F2, and Li2O? What kind of bonds are they? Solution: For C–O → |ΔEN| = |2 – 3| = 1.0 → This is a nonpolar covalent bond. For F–F → |ΔEN| = |4 – 4| = 0 → This is a nonpolar covalent bond. For Li2O → |ΔEN| = |1 – 3.5| = 2.5 → This is an ionic bond.

BOND ENERGY AND LENGTH TRENDS As we add more bonding (single, double, triple), the “glue” that binds two atoms to one another increases. This also means that the length of the bond between them decreases.

128

128

POLARIZABILITY As we travel down the periodic table, electron density gets more and more distorted. This results in a decrease in the ability to form strong bonds, hence the tendency for bond energies to go down.

For example, check out these bond strengths:

HF > HCl > HBr > HI.

Group VII HF 543

HCl 419 HBr 354

HI 287

Type of carbon-carbon bond Bond length (pm) Length trend Bond energy (kJ/mol) Energy trend C–C (single bond) 154 Length decreases,

bonding increases; inversely proportional.

348 Energy increases, bonding increases;

directly proportional. C=C (double bond) 134 612 C≡C (triple bond) 120 837

129

129

LECTURE 10: TURNING 2D STRUCTURES INTO 3D VSEPR MODELS TO INVESTIGATE POLARITY

The (Really) Big Picture

In our first lectures on bonding, we dealt in two-dimensions, our chemical structures being well-developed ionic and covalent 2D structures that were based upon simple concepts, like ΔEN and the stability of octet electronic configurations. Now we’ll become more sophisticated as we develop three-dimensional structures of covalently bonded compounds. We will use valence shell electron pair repulsion theory (VSEPR), and we’ll devise more satisfactory bonding orbitals based on valence bond theory (VB). Then we learn about a sophisticated mathematical model for bonding called molecular orbital theory (MO).

A sneak peek of what’s to come – an example of NH3:

As a Lewis structure: With VSEPR theory: With VB theory:

In VSEPR theory, the first thing you look for is the number of regions of electron density. Regions of e--density can be either a nonbonding e--pair or a bonding e--pair around an atom. Note that NH3 has four regions, three of which are bonding and one that’s nonbonding.

130

130

REGIONS OF ELECTRON DENSITY

The number of regions of e--density tells us three important structural features:

• Electronic geometry • Example: NH3 with 4 electron-rich regions ! its electronic geometry is tetrahedral

• Bond angles • Example: NH3 with 4 electron-rich regions ! its bond angles are 109.5°

• Hybridization • Example: NH3 with 4 electron-rich regions ! its hybridization is sp3

We’ll be going over these structural features in the next few lectures. Examples:

SO2 Number of e--rich regions = 3 Electronic geometry = trigonal planar Bond angles = 120° Hybridization = sp2

XeF2 Number of e--rich regions = 5 Electronic geometry = trigonal pyramidal Bond angles = 90°, 120°, 180° Hybridization = sp3d

CO32-

Number of e--rich regions = 3 Electronic geometry = trigonal planar Bond angles = 120° Hybridization = sp2

SF6 Number of e--rich regions = 6 Electronic geometry = octahedral Bond angles = 90°, 180° Hybridization = sp3d2

131

131

IDENTIFYING POLAR AND NONPOLAR COMPOUNDS USING VECTOR ALGEBRA And now a time-out for a bit of vector math so we can learn about another layer of chemical information: whether a compound is polar or nonpolar.

This bit of high school math will be used to add ΔEN vectors together. Why? – Because, by definition:

In a nonpolar molecule ! ΣΔEN = 0 In a polar molecule ! ΣΔEN ≠ 0

Example – CO2:

Find the ΔEN of each bond:

EN of C = 2.5 and EN of O = 3.5 → ΔEN = 3.5 – 2.5 = 1.0 Then sum the two bonds: ΣΔEN = 1.0 + (–1.0) = 0

Since ΣΔEN = 0, we know CO2 is nonpolar. Example – NH3:

The vectors do not cancel out, so ΣΔEN ≠ 0. Therefore, NH3 is polar.

132

132

A FAMOUS EXAMPLE – VSEPR SHOWS THAT H2O IS POLAR

As we know, the Lewis structure of H2O looks like this:

Since there are two O–H vectors going in opposite directions, it seems clear that ΣΔEN = 0, indicating that H2O is a nonpolar molecule. But showing H2O in three dimensions using VSEPR, instead of in two dimensions as a Lewis structure, changes the story:

It’s now clear that H2O is a polar molecule. We will later see that this fact has profound consequences on the chemistry of water solutions.

MORE EXAMPLES: SUMMING ΔEN VECTORS Let’s take our previous definitions of polar and nonpolar compounds and make them a bit more interesting:

In a nonpolar molecule: ΣΔEN = 0 or non-symmetrical. In a polar molecule: ΣΔEN ≠ 0 or symmetrical.

Here are some examples:

133

133

LECTURE 11: AN INTRODUCTION TO VB AND VSEPR THEORY Now that we can draw 2-dimensional structures using Lewis dot structures, let’s move on to doing that in 3 dimensions. How do we do that? By using valence shell electron pair repulsion (VSEPR) theory. VSEPR says that, around any central atom, electron-rich regions tend to move and stay as far away from each other as possible. But what are electron-rich regions?

There are 5 “foundational” shapes of VSEPR to consider:

134

134

AN EXCEPTION TO SIMPLE BOND ANGLES Bonded and unbonded e-pairs are not the same thing. Bonded e-pairs are constrained by two nuclei, so they take up less space (because they’re held more tightly and closely to the nuclei) than if they were constrained by a single nucleus:

A GLIMPSE OF THE NEXT THREE LECTURES

As we finish up our unit on bonding, we’ll be learning about some pretty advanced ideas and theories related to (guess what!) chemical bonds. A word of warning: it’s going to be important that you keep these concepts straight; they’re easy to mix up, and that will not make learning this already-tough stuff any easier. So pay close attention, and follow the sequence. Some names sound confusingly similar but are not the same, such as “molecular geometries” vs. “molecular orbitals” vs. “molecular orbital theory.” Of course they’re related, but as we teach them, you’ll want to keep them in their own compartments; you’ll see what I mean when we get there, but you should take my word for it now. To start you off, here are the main topics gone through in each of the next three lectures: This lecture’s (#11) overarching topic: VSEPR. Subtopics – electronic geometries, molecular geometries (pretty hard), multiple central atoms (pretty easy). Next lecture’s (#12) overarching topic: valence bond (VB) theory. Subtopics – atomic orbits, molecular orbits (really hard), hybrid orbits (kind of hard). Next-next lecture’s (#13) overarching topic: molecular orbital (MO) theory.

135

135

MOLECULAR GEOMETRIES Before we actually get started here, here’s a tip: the next few pages are about both electronic geometries and molecular geometries. Electronic geometries are the simpler, “let’s not put that much thought into it” version of molecular geometries (but we’re learning both, so obviously we are going to put that much thought into it). We’ll be talking a lot about both of them for the rest of the chapter, so be sure to keep them straight. At the beginning of this lecture, we looked briefly at e-rich regions. Now let’s think more deeply about them. Check out what happens when you distinguish bonded (B) e-pairs from unbonded (U) e-pairs:

Keeping this in mind, consider three compounds, all with 4 e-rich regions (and therefore tetrahedral):

136

136

How are they different? – In their numbers of B and U electrons: As you can see, paying attention to whether an e-pair is B or U makes a huge difference when it comes to assigning a molecule’s shape. And to put it very generally, that’s pretty much the difference between electronic geometries and molecular geometries: electronic geometries are based upon the number of e-rich regions, with no regard for B vs. U pairs; molecular geometries, on the other hand, are based upon the number of e-rich regions, and whether each e-pair is B or U. There are only 5 electronic geometries; they’re the “foundational” VSEPR shapes that we looked at in the beginning of this lecture. There are 15 molecular geometries (5 of which are the electronic geometries), but fortunately they have shapes that make sense. Let’s take a look at them before going into detail:

137

137

Each line extending from the center atom is an e-rich region. Notice that each e-rich region has either a dark lobe or a light circle at the end of it. The lobes represent unbonded e-pairs, and the circles represent bonded e-pairs. Most molecules have at least one unbonded pair, but not all of them do – the 5 that have only bonded electrons are the electronic geometry shapes. Those are the simplest of the molecular geometries. Also, even molecular geometries with U’s are assigned an electronic geometry; it’s always the one with the same number of e-rich regions. Unbonded pairs in a molecule obviously play a major role in determining its shape, but if you looked at a molecule, you wouldn’t see the unbonded pairs – just atoms held together by bonding electrons. So the point of molecular geometry is to let you see a molecule’s shape after its unbonded pairs have faded out of the image, because that’s what you’d really see if you looked at it (as opposed to a frumpy thing with funny lobes). The next two pages go over the molecular geometries you will need to know. Notes about these tables:

• The notation “AB#U#” is what we use to tell the number of bonded (B) and unbonded (U) electron pairs in the molecule. So a trigonal pyramidal molecule is AB3U, and a T-shape can either be AB3U2 or AB3U3.

• The far right column is for hybridization. You will learn about this in the next lecture, so don’t think you should already know it; it was just included for future convenience.

138

138

# of e-rich regions Electronic geometry AB#U# Shape Molecular geometry Bond

angle Hybridization

2 Linear AB2 Linear 180° sp

3

Trigonal planar

AB3

Trigonal planar 120°

sp2

3 AB2U

Bent < 120°

4

Tetrahedral

AB4

Tetrahedral 109.5°

sp3 4 AB3U

Trigonal pyramidal < 109.5°

4 AB2U2

Bent < 109.5°

5

Square bipyramidal

AB5

Trigonal bipyramidal 90°, 120°, 180°

dsp3

5 AB4U

See-saw 90°, 120°, 180°

5 AB3U2

T-shaped 90°, 180°

5 AB2U3

Linear 180°

139

139

Example: What are the electronic and molecular geometries for ICl2

-? Note: ICl2

- is like I3- and has 22 available electrons.

Solution:

# of e-rich regions Electronic geometry AB#U# Shape Molecular geometry Bond angle Hybridization

6

Octahedral

AB6

Octahedral 90°, 180°

d2sp3

6 AB5U

Square pyramidal 90°, 180°

6 AB4U2

Square planar 90°

6 AB3U3

T-shaped 90°, 180°

6 AB2U4

Linear 180°

140

140

MULTIPLE CENTRAL ATOMS: MUCH SIMPLER THAN YOU MIGHT THINK Even for molecules with more than one central atom, we can assign molecular geometries just as we would for simpler ones. We just look at each central atom individually and count the number of e-rich regions around it. So each central atom gets its own molecular geometry. For example, ethane has 2 central atoms (see it in the left column of the table below). Each of them has 4 electron-rich regions, so they’re both tetrahedral. This also applies to large organic molecules, like:

• Tetrahedral C’s → 1, 2 • Trigonal C’s → 3, 4, 5, 6

Here are 3 famous molecules with multiple central atoms. You must know these COLD. Ethane – C2H6 Ethylene – C2H4 Acetylene – C2H2

Tetrahedral Trigonal planar Linear

109.5° 120° 180° sp3 sp2 sp Sigma bonds: 7 Pi bonds: 0

Sigma bonds: 5 Pi bonds: 1

Sigma bonds: 2 Pi bonds: 2

141

141

LECTURE 12: USING VB THEORY TO MAKE MO’S FROM AO’S What is a bond? An overlap of electron-density from 2 atomic orbits (AO’s). Remember atomic orbits from our electronic configuration days?

We overlap two atomic orbits to make a valence bond (VB), also called a molecular orbital (MO):

So what are the types of molecular orbits? There are two: sigma and pi.

Sigma bonds (σ bonds) occur when the e-density is on the axis between atoms’ nuclei. (Figure on the left.) Pi bonds (π bonds) occur when the e-density is off the axis between atoms’ nuclei. (Figure on the right.)

142

142

Let’s make some sigma and pi bonds by mixing s and p atomic orbitals. For sigma bonds, we need to put the e-density on the axis between nuclei, and to do that, we just need our AO’s to meet (and overlap) in the middle. Using s and p orbitals, we can make sigma bonds with three combinations:

Notice that an s orbit is spherical, there is no way for it to make a bond that isn’t on the axis. That means s orbitals always form sigma bonds, so MO’s involving an s orbital will always be sigma bonds. Now onto learning about pi bonds by looking at the three orientations of p orbitals:

When a p orbital is involved in an MO, one of its orientations will form a σ bond, and the other two will form π bonds. There are three cases in which there are always π bonds:

1. Double bonds 2. Triple bonds 3. Resonance

Here is an example of a π bond in the a double bond of CH2O:

143

143

To summarize all of that: The atomic orbits:

• s orbitals → 1 orientation; sphere shape. • p orbitals → come in sets of 3, each with a different orientation (along x-, y-, or z-axis); lobe shapes.

The molecular orbits: • Sigma (σ) bonds → e-densities are on the same axis as nuclei, so bond occurs between them.

• Form anytime an s orbit is involved → s+s, s+p. • Of the 3 p orbits, one will form a σ bond.

• Pi (π) bonds → e-densities are off axis of nuclei, so bond appears to hover above and below them. • Of the 3 p orbits, two will form π bonds. • At least one π bond will always form in double bonds, triple bonds, or resonance.

And that’s almost it. Except we need to consider the case where AO’s become hybrid orbitals (which are types of AO’s). AO’s exist for all electrons of an atom, but hybrid orbitals are the version specific to valence electrons. So instead of the valence e’s being secluded in their original orbitals, all of their orbitals get (metaphorically) thrown into a blender. Ultimately, each valence electron ends up with a new “smoothie” orbital, made with equal amounts of all original orbitals thrown into it. The types of hybrid orbits are:

sp sp2 sp3 dsp3 d2sp3 And the specific type of hybrid orbit formed depends on which valence AO’s were the atom’s available ingredients. Note, happily, that every hybrid orbit has s character, which means they all have some e-density along the axis, so:

Every hybrid orbit that forms a valence bond with another atomic orbit makes a sigma, σ, bond. Plus, the fact that all hybrid orbitals have some s character means they all look the same:

This is how you can picture every hybrid orbit… Like a pregnant p.

144

144

Now that we know about hybrid orbits, let’s looks at the ways we can combine AO’s to form a sigma bond:

3 EXAMPLES OF ASSIGNING MO’S Let’s practice identifying the sigma and pi bonds – and the AO’s that formed them – in C2H6, C2H4, and C2H2.

Example – C2H6:

We already know how to figure out a molecule’s structure, and we’ll use that to determine its MO’s. In C2H6’s case, there are no double bonds, triple bonds, or resonance. C’s 4 valence electrons – each in an sp3 hybrid orbital – will bond with H’s electrons in their 1s orbitals:

• All bonds are sigma, so there is a total of 7 σ bonds → 6 are C-H, and 1 is C-C. • The C-H bonds are σ, made of electrons from H’s 1s orbit and C’s sp3 hybrid orbit → Notated: • The C-C bond is σ, made of e’s from each C’s sp3 hybrid orbits →

σ 1s-sp3 σ sp3-sp3

145

145

Example – C2H4:

After we’ve figured out the structure of C2H4, we see that there is a double bond present:

• The C-H bonds are σ, made of e’s from H’s 1s orbit and C’s sp2 hybrid orbit → • One of the C-C bonds is σ, made of e’s from each C’s sp2 hybrid orbit → • The other C-C bond is π, made of e’s from each C’s leftover p orbit →

Example – C2H2:

Once we know the structure of C2H2, we see there is a triple bond present in the molecule:

• Both C-H bonds are σ, made from H’s 1s orbit and C’s sp hybrid orbit →

• A C-C bond is σ, made from C’s sp hybrid orbits → • The other two C-C bonds are π, made of e’s from the two

leftover p orbits of each C →

σ 1s-sp σ sp-sp

π 2p-2p

π 2p-2p

σ 1s-sp2 σ sp2-sp2

146

146

VB THEORY EXPLAINS HYBRIDS At this point we’ve learned some stuff about hybrids, but we don’t yet know why they form in the first place. Unfortunately, VB (valence bond) theory is needed to explain that. If we didn’t know about hybrids, we would think that the molecular orbits in CH4 were formed like this:

That would mean that CH4 has 4 sigma bonds. One of the H’s would use its 1s electron to bond with the C’s 2s electron, and since the C only has a single 2s electron and which has already been put in a bond with H, the remaining 3 H’s must bond with C’s three 2p electrons.

So there should be 2 different types of bonds → one bond σ 1s-2s and the other three σ 1s-2p. With this thinking, spectroscopy should show 2 peaks, right? But it doesn’t! It shows a single peak:

147

147

So the 4 bonds must be the same – they must have made all of their orbitals equivalent:

HOW MANY KINDS OF HYBRIDS ORBITALS ARE THERE? For VSEPR, we learned about determining molecules’ shapes based on the number of e-rich regions, and the same idea applies to determining a molecule’s hybrid orbitals – the number of e-rich regions tells you the type (and number) of hybrid orbitals.

148

148

And in context of electronic geometries:

Example:

What is the hybridization of BeH2? Solution:

First step is to create Lewis structure (shown to the right). The 2 e-rich regions tell us that there are 2 hybrid orbits, which must have been made with 2 AO’s: One s orbit + one p orbit → gives us two sp hybrid orbits → BeH2 has sp hybridization.

Example: What is the hybridization for SF6?

Solution:

The Lewis structure (to the right) shows 6 e-rich regions, so SF6 has 6 hybrids, made from 6 AO’s: One s orbit + three p orbits + two d orbits → makes six d2sp3 hybrid orbits → it has d2sp3 hybridization.

149

149

LECTURE 13: MOLECULAR ORBITAL THEORY

Chapter Summary

You’ve probably been learning about chemical bonds for years, but has anyone ever explained to you why bonds come to happen in the first place? Well, we’ll learn about that in this lecture, by using molecular orbital (MO) theory to see the advantages and disadvantages to certain chemical bonds. Whereas VSEPR and VB focus only on the central atom (and can therefore not explain resonance), MO theory will let us understand more about what a molecular orbit is. MO theory will allow us to:

• Predict the existence of bonds. • Rank bond energy and length. • Predict paramagnetism. • Explain resonance.

MO theory is complex, so we will look only at the simple case of diatomic molecules such as H2, O2, N2, CO, and HF. Recall that bonds are formed from overlap of electron density from two atomic orbitals. The bond that is formed is a molecular orbital and to this point, we have simply called them σ and π bonds using VB theory.

In contrast, MO theory lets us view bonding in a more sophisticated manner by telling us that when atomic orbits overlap, they can either form an antibond or a bond:

150

150

Antibonding occurs if the MO formed is higher in energy than the individual AO’s. Bonding occurs if the MO formed is lower in energy than the individual AO’s.

We can draw out these antibonding and bonding situations by laying out AO’s (which we already know) in a pattern: Notice that as you go up the diagram, the energy increases. In these diagrams, you’ll notice that almost every other MO line has an asterisk (*) next to its “σ” or “π” label. This is used to differentiate bonding from antibonding MO’s – an asterisk indicates antibonding.

THE TWO SIMPLEST MO THEORY EXAMPLES Let’s check out the MO theory diagrams are for H2 and He2. In the case of H2, we get the figure to the right:

Because the MO formed is lower in energy than the individual AO’s, this bond is favorable, and H2 forms. And the diagram for a MO of two helium atoms is to the right:

Unlike in H2, the bond and antibond cancel each other out, so He2 doesn’t form.

151

151

BOND ORDER

A simple way to determine from MO theory whether a bond forms is by calculating bond order (b.o.), which we do with the following equation:

To put it simply, a bond order of 0 means that no bond will form, and a bond order greater than 0 means that a bond will form. More specifically:

• If b.o. = 1 → there is a single bond. • If b.o. = 2 → there is a double bond. • If b.o. = 3 → there is a triple bond.

Now we can apply this to the above examples of H2 and He2:

• H2 → b.o. = (2 – 0) / 2 = 1 → H2 has a single bond. • He2 → b.o. = (2 – 2) / 2 = 0 → He2 does not exist.

BOND ENERGIES AND LENGTHS Bond order can also tell us about bond length and bond energy ranking – b.o. has an indirect relationship to bond length and a direct relationship to bond energy. Here is an example ranking:

In the case above, N2’s bond has both the highest energy and the shortest length, because its b.o. is the largest.

152

152

PARAMAGNETISM It turns out that magnets form from unpaired electrons. This means that any time we draw an AO and MO and see a single, unpaired electron, we are actually looking at a magnet, and we call it paramagnetism. This means that molecules in which the electrons do not pair up – like B2 – are paramagnetic. In fact, O2, which has unpaired electrons in the antibonding π MO’s, is experimentally observed to be a magnet.

CREATING MO’S FOR ALL HOMONUCLEAR DIATOMS A homonuclear diatom is exactly what it sounds like – a molecule made up of two of the same atoms. We’re going to learn how to fill in the bonding and antibonding orbitals for all diatoms of the second-row elements. Since this means we’ll be working with another energy level (n = 2), we have to expand the MO theory diagram:

153

153

Some things to notice about that extended MO energy level chart: 1. For every AO, there is an MO. 2. For every bonding MO, there is an antibonding MO. 3. The six 2p AO’s make 6 MO’s – two σ bonds (one bonding, one antibonding), and four π bonds (one bonding and

one antibonding for each of the three axes). 4. Just as in VB theory, we see there’s one σ bond for every two π bonds because there are two axes for π bonds.

There is one other complication for the second-row diatomic atoms: for the larger ones, the order of MO’s change at the top of the energy chart. So for diatoms H2 through N2, the MO’s are ordered as shown in the left half of the diagram below, and for diatoms O2

through Ne2, the MO’s are order as shown in the right half:

This order looks like a frowny face.

This order looks like a smiley face.

154

154

THE GREATEST TABLE EVER Presented below is a table of MO data for all the second-row homonuclear diatomic molecules. The table presents the order in which the MO’s fill and also provides experimental data that confirms how the bond order for each compound corresponds to concepts like magnetism, bond length, and bond energy. It also shows the reordering of MO’s for the larger diatomic compounds. Learn this table cold and you’ll be able to handle 90% of MO theory in this class.

HETERONUCLEAR DIATOMS We’ll stick to pretty simple cases of MO’s in this class, but there are a few more complex ones we’re interested in, like heteronuclear diatomic compounds. We won’t go into too much depth, but there is something to be learned from simply following the rules for homonuclear cases, adding the total number of AO electrons, and then filling the MO’s. For example, CO’s 14 total electrons can have MO’s formed by following Aufbau, Pauli, Hund, and the MO theory rules above. The energy levels will not be correct, but concepts like bond order and paramagnetism will still apply.

155

155